k^2=103

Solution for k^2=103 equation:

k^2=103

We move all terms to the left:

k^2-(103)=0

a = 1; b = 0; c = -103;
Δ = b2-4ac
Δ = 02-4·1·(-103)
Δ = 412
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{412}=\sqrt{4*103}=\sqrt{4}*\sqrt{103}=2\sqrt{103}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{103}}{2*1}=\frac{0-2\sqrt{103}}{2}
=-\frac{2\sqrt{103}}{2}
=-\sqrt{103}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{103}}{2*1}=\frac{0+2\sqrt{103}}{2}
=\frac{2\sqrt{103}}{2}
=\sqrt{103}
$